Carrying a Heavy Bag Across a Room Is Zero Work

Because the force is upward and the movement is forward, and work counts only the movement that happens along the force.

A coolie balancing a trunk on their head pushes up on it. Walking along a platform moves it sideways. Not a single centimetre of that motion is in the direction of the upward push, so the work done on the trunk is exactly zero.

That sounds absurd until you notice physics is not measuring tiredness. It is measuring transfer of energy, and none is being transferred to the trunk while it travels level. This page covers the first part of the ICSE Class 8 Physics chapter on energy: work, the cases where it vanishes, and power.

In three distinct situations — and each is a favourite exam question.

1. There is no displacement. Push a heavy almirah that will not budge. The force may be $300\text{N}$, but $d=0$:

$$W=300\times 0=0\text{J}$$

The same goes for pushing a wall. You grow tired because your muscles are using energy internally, but no work is done on the wall.

2. The displacement is perpendicular to the force. The coolie on the platform, or a student walking with a school bag: the supporting force is vertical, the motion horizontal. No part of the displacement lies along the force, so the work is zero.

A planet circling the Sun is the same case — the pull is towards the centre while the motion is around it.

3. There is no force. A ball moving on a perfectly smooth, level surface keeps going with no force needed, so no work is being done on it.

**Contrast these with the cases where work is done. Lifting the bag off the ground does work, because now the motion is upward, along the force. So the coolie does work picking the trunk up and putting it down**, and none at all in between.

That boundary — same trunk, same person, work in one phase and zero in the next — is the clearest test of whether you have understood the definition rather than memorised it.

Power is the rate of doing work — the work done in each second:

$$P=\frac{W}{t}$$

Its SI unit is the watt, $\text{W}$, and

$$1\text{W}=1\text{J/s}$$

Worked example. The $150\text{J}$ of work from lifting the bag, done in $3\text{s}$:

$$P=\frac{150}{3}=50\text{W}$$

Now lift the same bag the same height in $1.5\text{s}$:

$$P=\frac{150}{1.5}=100\text{W}$$

The work is identical — same force, same height — but the power has doubled, because the same energy was transferred in half the time.

This is the distinction that gets tested. Two people carry identical loads up the same staircase; the one who runs up does the same work but develops more power. Work measures how much energy moved; power measures how fast.

Reading it backwards. Rearranging gives two more useful forms:

$$W=P\times t,t=\frac{W}{P}$$

So a $40\text{W}$ bulb left on for $60\text{s}$ consumes

$$W=40\times 60=2400\text{J}$$

of energy — which is why the wattage printed on an appliance tells you its rate of energy use, not its total.

Two conversions cover every numerical in this chapter:

$$1\text{kW}=1000\text{W},1\text{hp}=746\text{W}$$

The horsepower is an older unit still printed on motors and pumps sold in India, which is why the syllabus keeps it.

Worked example — a water pump. A pump raises $200\text{kg}$ of water to a tank $10\text{m}$ above it in $40\text{s}$. Find its power in watt, kilowatt and horsepower.

First the work, using $W=mgh$:

$$W=200\times 10\times 10=20000\text{J}$$

Then the power:

$$P=\frac{20000}{40}=500\text{W}$$

In kilowatt:

$$P=\frac{500}{1000}=0.5\text{kW}$$

In horsepower:

$$P=\frac{500}{746}\approx 0.67\text{hp}$$

Worked example — the other direction. A motor is marked $2\text{hp}$. Its power in SI units is

$$2\times 746=1492\text{W}\approx 1.49\text{kW}$$

A three-step habit for every numerical. Find the work first, then divide by the time, then convert the unit if asked. Trying to reach horsepower in one line is how sign and factor errors creep in.

And notice the pump problem never needed the tank's size or the pipe's width. Only mass, height and time entered the calculation — a reminder that $W=mgh$ counts the vertical rise and nothing else.